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**>> 7.5. Systems of Inequalities in Two Variables **

7.4. Systems of Nonlinear Equations in Two Variables

Thus far in this chapter we have discussed how to solve systems of linear equations. Now we shift to systems of nonlinear equations. These are considerably more difficult to deal with in general. However, the standard technique of substitution may be used for these systems. Occasionally, other tricks can be used as well. For simplicity, we will restrict our discussion to systems of just two nonlinear equations.

**Example 1:** Solve the following system of nonlinear equations:

x^{2} + y^{2} = 25

x + 3y = 15

**Solution: **We first solve the second equation for x in terms of y, obtaining x = 15 - 3y. Substituting this equation into the first equation yields (15 - 3y)^{2} + y^{2} = 25. Expanding the square on the left yields 225 - 90y + 9y^{2} + y^{2} = 25, which simplifies to 225 - 90y + 10y^{2} = 25. Subtracting 25 from each side, we obtain the following quadratic equation in y:

200 - 90y + 10y^{2} = 0.

Note that we can simplify this equation by dividing both sides by 10, obtaining

20 - 9y + y^{2} = 0.

By trial and error or by means of the quadratic formula, we find the solutions y = 4 and y = 5. For each of these solutions we must also solve for x. The second of our original equation can be used to solve explicitly for x in terms of y; we have x = 15 - 3y. Thus, when y = 4 we have x = 15 - (3)(4) = 3 and when y = 5 we have x = 15 - (3)(5) = 0, yielding the two solutions x = 3; y = 4 and x = 5; y = 0.

Note that this system of nonlinear has exactly two distinct solutions, something that can never happen with linear equations. In general, a system consisting of two polynomial equations in two variables, one of degree m and the other of degree n, can have as many as mn solutions.

**Example 2:**

Solve the following system of nonlinear equations:

x + y = 1

1/x + 1/y = 4

**Solution: **First we multiply the second equation by xy, obtaining y + x = 4xy. By virtue of the first equation, this implies 4xy = 1, which in turn implies y = 1 / 4x. Plugging this into the second equation yields 1/x + 4x = 4. Multiplying this equation by x yields 1 + 4x^{2} = 4x, which simplifies to 4x^{2} - 4x + 1 = 0. But the left side is a perfect square, namely the square of 2x - 1, whence (2x - 1)^{2} = 0, implying 2x - 1 = 0 or x = 1/2. Plugging this value of x back into the first equation, we find y = 1/2. Thus, the unique solution is x = 1/2; y = 1/2.

**Example 3:**

Solve the following system of nonlinear equations:

x^{2} + y^{2} = 4

xy = 1

**Solution: **We can be a bit clever here. Note that by adding or subtracting the second equation from the first, we obtain a perfect square on the left. Thus we find

x^{2} + 2xy + y^{2} = (x + y)^{2} = 6

x^{2} - 2xy + y^{2} = (x - y)^{2} = 2

whence we have the following four systems of linear equations in x and y:

x + y = ±√6

x - y = ±√2

Adding these equations and dividing both sides by 2 yields x = (±√6 ± √2) / 2. Similarly, subtracting the second equation from the first yields y = (±√6 ± √2) / 2, where the signs of the first ± symbols are equal and thos of the second on are opposite. Thus, the four solutions are as follows:

- x = (√6 + √2) / 2; y = (√6 - √2) / 2
- x = (√6 - √2) / 2; y = (√6 + √2) / 2
- x = (-√6 + √2) / 2; y = (-√6 - √2) / 2
- x = (-√6 - √2) / 2; y = (-√6 + √2) / 2

**Example 4:** Solve the following system of nonlinear equations:

x^{2} + y^{2} = 1

x^{4} + y^{4} = 1

**Solution: **We may use the first equation to solve for y^{2} in terms of x, obtaining y^{2} = 1 - x^{2}. Plugging this into the second equation yields x^{4} + (1 - x^{2})^{2} = 1, which when expanded yields x^{4} + 1 - 2x^{2} + x^{4} = 1, which simplifies to 2x^{4} - 2x^{2} = 0. Dividing this equation by 2 yields x^{4} - x^{2} = 0, which implies x = 0 or x^{2} = 1. Thus we have x = ±1 or x = 0. It is easy to see that x = ±1 implies y = 0 and x = 0 implies y = ±1. Thus we have found the following four solutions:

- x = 1; y = 0
- x = 0; y =1
- x = -1; y =0
- x = 0; y = -1

Occasionally, a system of nonlinear equations can be converted to a set of linear equations, as the following example shows:

**Example 5:** Solve the following system of nonlinear equations:

y^{5} = 4x^{3}

y^{8} = 8x^{5}

**Solution: **We may solve this equation by taking the base-2 logarithm of each equation, obtaining the system

5 log_{2}y ^{} = 3 log_{2}x + 2

8 log_{2}y ^{} = 5 log_{2}x + 3

Now if we let u = log_{2}x and v = log_{2}y, we get the following system of two linear equations in two variables.

3u - 5v = 2

5u - 8v = 3

But we know how to solve such systems from Section 7.1 (also from section P.9). Omitting the details, we obtain the solution u = v = 1, whence x = 2^{u} = 2 and y = 2^{v} = 2. Thus, the solution is x = y = 2.

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